# Is the function, $f(z)=iz\bar{z}$ analytic?

Question: Is the function, $$f(z)=iz\bar{z}$$ analytic?

My approach: We know that for any $$z\in\mathbb{C}$$, $$z\bar{z}=|z|^2.$$ Thus $$f(z)=i|z|^2, \forall z.$$ Now let $$z=x+iy\implies f(z)=f(x+iy)=i(x^2+y^2).$$ Thus we have $$u(x,y)=0$$ and $$v(x,y)=x^2+y^2$$, $$\forall x,y\in\mathbb{R}$$.

Now observe that both $$u$$ and $$v$$ are continuous functions in $$x$$ and $$y$$, $$\forall x,y\in\mathbb{R}.$$

Now $$\frac{\partial{u}}{\partial{x}}=0, \frac{\partial{u}}{\partial{y}}=0,\frac{\partial{v}}{\partial{x}}=2x,\frac{\partial{v}}{\partial{y}}=2y, \forall x,y\in\mathbb{R}.$$

Thus the functions $$\frac{\partial{u}}{\partial{x}},\frac{\partial{u}}{\partial{y}},\frac{\partial{v}}{\partial{x}},\frac{\partial{v}}{\partial{y}}$$ are continuous $$\forall x,y\in\mathbb{R}.$$

Now we move forward to analyze if the Cauchy-Riemann conditions are satisfied or not.

Observe that since $$\frac{\partial{u}}{\partial{x}}=0 \text{ and } \frac{\partial{v}}{\partial{y}}=2y, \text{ therefore } \frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}}\iff 2y=0\iff y=0.$$

Again since, $$\frac{\partial{u}}{\partial{y}}=0 \text{ and } -\frac{\partial{v}}{\partial{x}}=-2x, \text{ therefore } \frac{\partial{u}}{\partial{y}}=-\frac{\partial{v}}{\partial{x}}\iff -2x=0\iff x=0.$$

Thus the Cauchy-Riemann conditions are satisfied only at the point $$(x,y)=(0,0)$$, i.e., at the point $$z=0$$.

Thus we can conclude that $$f$$ is analytic only at the point $$z=0$$.

I have read in the book "Advanced Engineering Mathematics" by Erwin Kreyszig that:

A function $$f(z)$$ is said to be analytic at a point $$z=z_0$$ in a domain $$D$$ if $$f(z)$$ is analytic in a neighborhood of $$z_0$$.

But, here we see that $$f(z)$$ is analytic only at the point $$z=0$$, and not in any $$\delta$$ neighborhood of $$z=0$$. So, can we still conclude that $$f(z)$$ is analytic at the point $$z=0$$? And, obviously $$f(z)$$ is not a analytic function, since it is not differentiable at every point in some domain.

• No, as the definition says, a function cannot be analytic at a single point, though it can be differentiable at a single point. May 5, 2020 at 15:52
• $f(z)$ is complex differentiable at $z=0$, but since it is not differentiable on any open set containing the origin, it is not analytic anywhere May 5, 2020 at 15:52
• That $z\bar z$ is a red herring: take any real-valued function instead. May 5, 2020 at 16:33

If the Cauchy-Riemann equations were satisfied in an open set $$D$$, you could conclude that your function was analytic in $$D$$. But here the C-R equations are only satisfied at one point. That does not say your function is analytic at that point, it says your function is not analytic at all, because to be analytic at a point (according to the definition) it must be analytic in a neighbourhood of the point.
Cauchy-Riemann only tells you when $$f$$ is complex differentiable at $$0$$, not holomorphic at $$0$$.